暴力最高\(50\)吧，本地测试不太准跑得快的只得了\(10\)分，慢的却得了\(50\)分
暴力
#include <bits/stdc++.h>
#define pb push_back
#define ll long long
#define bs bitset<70>
#define speed() ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
using namespace std;
const int N = 1e6+5;
int n,a[N],m;
struct node
{mutable int v;bool operator < (const node& A)const{return v<A.v;}
};
multiset<node> s;
int tmp[N];
int main()
{speed();freopen("sample1.in","r",stdin);freopen("out.out","w",stdout);cin>>n>>m;for(int i=1;i<=n;i++){cin>>a[i];s.insert({a[i]});}ll ans=0;// cout<<"*******"<<endl;while(s.size()){ auto it=s.end();int mi=1e9;for(int j=1;j<=m;j++){it--;tmp[j]=(*it).v;mi=min(tmp[j],mi);if(it==s.begin())break;}s.erase(it,s.end());for(int j=1;j<=m;j++,it--){// cout<<tmp[j]<<" ";tmp[j]-=mi;if(tmp[j]>0)s.insert({tmp[j]});}// cout<<endl;ans+=mi;}cout<<ans<<endl;return 0;
}
其实这是一道性质题，发现了很显然，\(max(\frac{\sum a_i}{m},a[n])\)
点击查看代码
#include <bits/stdc++.h>
#define pb push_back
#define ll long long
#define bs bitset<70>
#define speed() ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
using namespace std;
const int N = 1e6+5;
ll n,a[N],m;
int main()
{speed();// freopen("sample1.in","r",stdin);// freopen("out.out","w",stdout);cin>>n>>m;ll sum=0,mx=0;for(int i=1;i<=n;i++){cin>>a[i];mx=max(mx,a[i]);sum+=a[i];}sort(a+1,a+1+n);cout<<max((ll)ceil(1.0*sum/m),mx);return 0;
}
T2 P9133 [THUPC 2023 初赛] 大富翁
我们可以把要求的转换为式子，设\(f(u,v)\)为\(u\)支配\(v\),题面有点歧义,不知道支配是算上过去和未来的，还是只算过去的
\[Ans=\sum_{x \in S}\sum_{y \in T}f(x,y)-\sum_{x \in S}\sum_{y \in T
}f(y,x)- \sum_{x \in S}w_x\]
\[=\sum_{x \in S}(\sum_{y \in T}f(x,y)+ \sum_{y \in S}f(x,y)
)-\sum_{x \in S}(\sum_{y \in T}f(y,x)+ \sum_{y \in S}f(y,x)
)- \sum_{x \in S}w_x\]
\[=\sum_{x \in S}sz_x
-\sum_{x \in S}dep_x
-\sum_{x \in S}w_x\]
\[=\sum_{x \in S}(sz_x - dep_x - w_x)
\]
点击查看代码
#include <bits/stdc++.h>
#define pb push_back
#define ll long long
#define bs bitset<70>
#define speed() ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
using namespace std;
const int N = 1e6+5;
int n,a[N],m,w[N],f[N],dep[N],sz[N];
vector <int> edge[N];
struct ca
{int w,id;
}q[N];
void dfs(int u)
{dep[u]=dep[f[u]]+1;sz[u]=1;for(auto to:edge[u]){dfs(to);sz[u]+=sz[to];}
}
bool cmp(ca a,ca b){return a.w>b.w;}
int main()
{speed();freopen("sample2.in","r",stdin);freopen("out.out","w",stdout);cin>>n;for(int i=1;i<=n;i++)cin>>w[i];for(int i=2;i<=n;i++)cin>>f[i],edge[f[i]].pb(i);ll ans=0;dfs(1);for(int i=1;i<=n;i++){q[i].id=i;q[i].w=sz[i]-dep[i]-w[i];}sort(q+1,q+1+n,cmp);for(int i=1;i<=n;i++){if(i&1)ans+=q[i].w;}cout<<ans;return 0;
}
